Finding the Second Derivative dx/dy Squared

This question asks for the value of the second derivative of x with respect to y, which is written as \(\dfrac{d^2x}{dy^2}\). This involves understanding how to change the dependent and independent variables in differentiation, specifically for second-order derivatives.

Step 1: Relationship Between First Derivatives

First, recall the relationship between the first derivative of x with respect to y and the first derivative of y with respect to x. If \(y = f(x)\), then:

• \(\dfrac{dx}{dy} = \dfrac{1}{\frac{dy}{dx}}\)
• This can also be written using negative exponents as: \(\dfrac{dx}{dy} = \left(\dfrac{dy}{dx}\right)^{-1}\)

Step 2: Differentiating to Find the Second Derivative

To find \(\dfrac{d^2x}{dy^2}\), we need to differentiate \(\dfrac{dx}{dy}\) with respect to y. So, we start with:

• \(\dfrac{d^2x}{dy^2} = \dfrac{d}{dy} \left( \dfrac{dx}{dy} \right)\)
• Substitute the expression for \(\dfrac{dx}{dy}\) from Step 1:
• \(\dfrac{d^2x}{dy^2} = \dfrac{d}{dy} \left[ \left(\dfrac{dy}{dx}\right)^{-1} \right]\)

Step 3: Applying the Chain Rule

In the expression \(\left(\dfrac{dy}{dx}\right)^{-1}\), the term \(\dfrac{dy}{dx}\) is a function of x. Since we are differentiating with respect to y, and x is implicitly a function of y, we must use the chain rule. The chain rule states:

• \(\dfrac{d}{dy}(g(x)) = \dfrac{d}{dx}(g(x)) \cdot \dfrac{dx}{dy}\)

Let \(g(x) = \left(\dfrac{dy}{dx}\right)^{-1}\). First, find the derivative of \(g(x)\) with respect to x:

• \(\dfrac{d}{dx} \left[ \left(\dfrac{dy}{dx}\right)^{-1} \right] = -1 \cdot \left(\dfrac{dy}{dx}\right)^{-1-1} \cdot \dfrac{d}{dx}\left(\dfrac{dy}{dx}\right)\)
• Using the power rule and the definition of the second derivative \(\dfrac{d^2y}{dx^2}\):
• \(\dfrac{d}{dx} \left[ \left(\dfrac{dy}{dx}\right)^{-1} \right] = - \left(\dfrac{dy}{dx}\right)^{-2} \cdot \dfrac{d^2y}{dx^2}\)

Now, apply the chain rule by multiplying this result by \(\dfrac{dx}{dy}\):

• \(\dfrac{d^2x}{dy^2} = \left[ - \left(\dfrac{dy}{dx}\right)^{-2} \cdot \dfrac{d^2y}{dx^2} \right] \cdot \dfrac{dx}{dy}\)

Step 4: Final Simplification

Substitute \(\dfrac{dx}{dy} = \left(\dfrac{dy}{dx}\right)^{-1}\) into the equation obtained in Step 3:

• \(\dfrac{d^2x}{dy^2} = - \left(\dfrac{dy}{dx}\right)^{-2} \cdot \dfrac{d^2y}{dx^2} \cdot \left(\dfrac{dy}{dx}\right)^{-1}\)

Combine the terms with the base \(\left(\dfrac{dy}{dx}\right)\) by adding their exponents:

• \(\dfrac{d^2x}{dy^2} = - \dfrac{d^2y}{dx^2} \left(\dfrac{dy}{dx}\right)^{-2 + (-1)}\)
• \(\dfrac{d^2x}{dy^2} = - \dfrac{d^2y}{dx^2} \left(\dfrac{dy}{dx}\right)^{-3}\)

This final expression matches the formula provided in option 4.